Django 无法将所有事件保存在数据库中

Question

我有一个可以获取所有事件详细信息的工作代码。但是我无法在数据库中保存完整的数据，只能保存最后一条记录

def incidents(request):
incidentsServicenow = IncidentsServicenow()
c = pysnow.Client(instance='', user='', password='')

# Define a resource, here we'll use the incident table API
incident = c.resource(api_path='/table/incident')
print('incident', incident)
# Query for incidents with state 1
response = incident.get()
print('response', response)
# Iterate over the result and print out `sys_id` of the matching records.
ticket = []
for record in response.all():
    data = {
        'number': record['number'],
        'description': record['description'],
        'short_description': record['short_description'],
        'state': record['state'],
    }
    #print(record['number'])
    incidentsServicenow.number = data['number']
    incidentsServicenow.title = data['short_description']
    incidentsServicenow.description = data['description']
    incidentsServicenow.state = data['state']
    #ticket.append(data)
    print("storing data")
    incidentsServicenow.save()
return HttpResponse(ticket, content_type='application/json')

我的模特是

class IncidentsServicenow(models.Model):
number = models.CharField(max_length=32)
title = models.CharField(max_length=160)
description = models.TextField()
state = models.CharField(max_length=20)

class Meta:
    managed = False
    db_table = 'incidents_servicenow'

我需要将所有记录保存在数据库中

Answer 1

您应该在循环中创建对象。从代码中，我可以看到您创建的 incidentsServicenow 对象是在循环外创建的。

  for record in response.all():
    data = {
        'number': record['number'],
        'description': record['description'],
        'short_description': record['short_description'],
        'state': record['state'],
    }
    #print(record['number'])
    incidentsServicenow = IncidentsServicenow()
    incidentsServicenow.number = data['number']
    incidentsServicenow.title = data['short_description']
    incidentsServicenow.description = data['description']
    incidentsServicenow.state = data['state']
    ticket.append(data)
    print("storing data")
    incidentsServicenow.save()
return HttpResponse(ticket, content_type='application/json')

或者你可以这样做

   for record in response.all():
    data = {
        'number': record['number'],
        'description': record['description'],
        'short_description': record['short_description'],
        'state': record['state'],
    }
    #print(record['number'])
    incidentsServicenow = IncidentsServicenow(number=data['number'],
title=data['short_description'],description=data['description'],state=data['state'])
    ticket.append(data)
    print("storing data")
    incidentsServicenow.save()
return HttpResponse(ticket, content_type='application/json')

Answer 2

在 for 循环中添加下行

incidentsServicenow = IncidentsServicenow()

Django 无法将所有事件保存在数据库中

django not able to save all incidents in database

django

django-database

django-datatable

django-database-functions